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Units and Measurements question

2025 · 22 Jan · Shift 1 · Q63
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Units and Measurements question

2025 · 22 Jan · Shift 1 · Q63

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
Given below are two statements : Statement I: In a vernier callipers, one vernier scale division is always smaller than one main scale division. Statement II : The vernier constant is given by one main scale division multiplied by the number of vernier scale divisions. In the light of the above statements, choose the correct answer from the options given below.
  1. A
    Both Statement I and Statement II are false
  2. B
    Statement I is true but Statement II is false
  3. C
    Both Statement I and Statement II are true
  4. D
    Statement I is false but Statement II is true
View written solutionFree

Correct answer: B

  1. Check Statement I

    In a standard vernier calipers, n VSD=(n−1) MSDn\text{ VSD} = (n-1)\text{ MSD}n VSD=(n−1) MSD for a direct vernier.

    Therefore, 1 VSD=n−1n MSD1\text{ VSD} = \frac{n-1}{n}\text{ MSD}1 VSD=nn−1​ MSD

    Since n−1n<1,\frac{n-1}{n} < 1,nn−1​<1, we get 1 VSD<1 MSD1\text{ VSD} < 1\text{ MSD}1 VSD<1 MSD

    So, Statement I is true.

  2. Check Statement II

    Vernier constant (least count) is defined as: Vernier constant=1 MSD−1 VSD\text{Vernier constant} = 1\text{ MSD} - 1\text{ VSD}Vernier constant=1 MSD−1 VSD for a direct vernier.

    It is not equal to 1 MSD×number of VSD1\text{ MSD} \times \text{number of VSD}1 MSD×number of VSD

    In fact, if nnn vernier divisions coincide with (n−1)(n-1)(n−1) main scale divisions, then VC=1 MSD−1 VSD=1n MSD\text{VC} = 1\text{ MSD} - 1\text{ VSD} = \frac{1}{n}\text{ MSD}VC=1 MSD−1 VSD=n1​ MSD

    Hence, Statement II is false.

  3. Conclusion

    • Statement I: True
    • Statement II: False

    Therefore, the correct option is: B\boxed{\text{B}}B​

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